《6 Sigma项目运作实例》

Multiple t-tests – Adequate?
An experimenter could run 2-sample t-tests against every
combination of means
1 vs. 2, 1 vs. 3, 1 vs. 4, 1 vs. 5
2 vs. 3, 2 vs. 4, 2 vs. 5
3 vs. 4, 3 vs. 5
4 vs. 5
Why would this not be a good idea?
Obviously, it is tedious and cumbersome
What about alpha risk?
Each t-test has a risk of false rejection ()
4 . 0
10 ^ 95 . 0 1
=
=total
Remember, the probability of being
successful in a multi-step process is the
product of the individual step probabilities.
Each of the 10 combinations of tests would
need to be correct in order for the entire
analysis is correct, that is, 0.95 x 0.95 x … x
.095 for each of the ten processes. The total
alpha risk is one minus the probability of
being correct, or in this case ~ 0.4. That is,
approximately 40% of the time in a five
treatment analysis like this, we would
incorrectly identify a difference that does not,
in truth, exist.
Solution – Analysis of Variance
ANOVA is really an extension (generalization) of the
2-sample t-test
ANOVA is a method of detecting difference between
multiple means of samples
Why is it called Analysis of Variance?
ANOVA is the mathematics behind the intuitive evaluation
ANOVA compares/analyzes variances
Variance within a group
Variance between groups
ANOVA is also the basis for analyzing
DOE, which will be covered in Week 3.

Minitab ANOVA Results – Case 1
One-way ANOVA: TmtA, TmtB, TmtC, TmtD, TmtCntrl1
Analysis of Variance
Source DF SS MS F P
Factor 4 24.65720 6.16430 643.60 0.000
Error 45 0.43100 0.00958
Total 49 25.08820
Individual 95% CIs For Mean
Based on Pooled StDev
Level N Mean StDev ----------+---------+---------+------TmtA
10 6.5500 0.0850 (*)
TmtB 10 6.9500 0.0850 (*)
TmtC 10 5.9900 0.1101 (*)
TmtD 10 7.5200 0.1033 (*)
TmtCntrl 10 5.5200 0.1033 (*)
----------+---------+---------+------Pooled
StDev = 0.0979 6.00 6.60 7.20
What do you
think this
means?
Minitab ANOVA Results – Case 2
One-way ANOVA: Tmt1, Tmt2, Tmt3, Tmt4, TmtCntrl2
Analysis of Variance
Source DF SS MS F P
Factor 4 31.51 7.88 3.89 0.009
Error 45 91.25 2.03
Total 49 122.76
Individual 95% CIs For Mean
Based on Pooled StDev
Level N Mean StDev --+---------+---------+---------+----Tmt1
10 6.250 1.457 (------*-------)
Tmt2 10 7.260 1.561 (-------*------)
Tmt3 10 6.240 0.729 (-------*-------)
Tmt4 10 7.710 1.412 (------*-------)
TmtCntrl 10 5.490 1.748 (-------*------)
--+---------+---------+---------+----Pooled
StDev = 1.424 4.8 6.0 7.2 8.4
What do you
think this
means?
Analysis of Variance – General Recipe

1. State the practical

problem

1. State the null

hypothesis

1. State the alternate

hypothesis

1. Do the model

assumptions hold?

1. Construct the

Analysis of Variance
Table

1. Do the assumptions for

the errors hold (residual
analysis)?

1. Interpret the p-value (or

the F-statistic) for the
factor effect (p < α )

1. Calculate %SSfor the

factor and error terms

1. Translate the

conclusion into practical
terms
This is the general recipe for hypothesis
testing. The tests differ in the appropriate
statistics and appropriate distributions.
偶然性检验:
Contingency Tables & Chi-Square Tests
What Is a Contingency Table?
Pg 7 ?April 01, Breakthrough Management Group. Unpublished proprietary work available only under license. All rights reserved. April 10, 2001
A contingency table is
A method used to analyze data in a two-way classification
A method for analyzing data that are usually discrete,
attribute data (count-type)
A tool for testing the relationship between two sources of
variation
Are the sources independent? Are the sources related?
Analogous to finding interactions in regression and ANOVA
analyses
Hypotheses:
H o : Factor A is independent of Factor B
H a : Factor A is not independent of Factor B
During the analysis of sources of
variation, it is often required to determine if
two factors that contribute to the variation are
independent of each other. Consider a gas-phase
chemical reactor vessel. The Black
Belt may find that both temperature and
pressure are critical factors contributing to
variance in the product yield. Given the
nature of a gas reaction and Charles?Law
relating temperature and pressure, it is very
likely that pressure is a function of
temperature and not an independent factor.
A contingency table analysis will answer
the question of independence.
This type of analysis is called a
contingency table because the alternate
hypothesis is that one factor is is not
independent (is contingent) on another.
Contingency Table Terminology
The analysis examples, so far, have all be univariate
Univariate – having one variable; measuring only one
characteristic in each sampling unit
Examples: Blood pressure reading on a patient, number of
crashes at an intersection in a month
Bivariate or Multivariate
Measurements taken on 2 or more characteristics per
sampling unit.
Examples: Surveys that correlate answers to demographic
data, a study of voting habits by education level.
Contingency tables are concerned with
bivariate or multivariate data.
A Contingency Table Example
The board of regents at a university would like to determine
if two variables are independent: employee classification
(staff, faculty, administrator) and whether a single union
should be the sole collective bargaining agent for employee
benefits.
A random sample of 200 employees is taken from
employee records and each employee is classified
according to both variables.
The data are in Excel worksheet CTables.xls and on the
following slide
Minitab Output
Expected counts are printed below observed counts
Favor NotFavor Undecide Total
1 30 1515 60
24.00 27.00 9.00
2 40 5010100
40.00 45.00 15.00
3 10 25540
16.00 18.00 6.00
Total 80 90 30 200
Chi-Sq = 1.500 + 5.333 + 4.000 +
0.000 + 0.556 + 1.667 +
2.250 + 2.722 + 0.167 = 18.194
DF = 4, P-Value = 0.001
Does this
look
familiar?
?
P-value<0.05
Reject Ho
Notice, once-again, that the Chi-Square
value form Minitab is the same as the one
calculated manually a few slides earlier.
Reject H0 if the p-value is below your
selected alpha-risk.
样品尺寸选择:
Why Sample Size Is Important:
A Class Exercise in Minitab
Start Minitab -- Select Calc > Random Data > Normal…
Generate two data sets
Sample1
μ= 0, σ = 1, n=10
Sample 2
μ= 0.5, σ = 1, n=10
Conduct a 2-sided,
2-sample
t-test on the data.
What do you conclude?
Repeat with n = 25,
n=50, n=75 &
n=100.
Record your results
for a later exercise.
Remember the Central Limit Theorem?
As the sample size increases, the standard
deviation of the sample means gets smaller. t-
tests are tests for comparing the means of
sample sizes. The smaller the standard
deviation, compared to the difference in the
mean, the greater the likelihood of detecting
the difference.
The Effect of Sample Size
As the graphs show, each distribution of means gets progressively more
narrow, until there is very little overlap between the two distributions.
This is a simple illustration of the effect of sample size on the the ability to
resolve differences between the means of two normal distributions. Each
of the other hypothesis tests also has a similar relationship to sample size.
The Confidence Interval for Means
As the equation shows, as n gets very
large, the width of the confidence interval will
be very small. At infinite sample sizes, the
width of the confidence interval is zero.
With enough samples, any difference will
be statistically significant. However, not
every statistically significant difference is of
practical importance.
Learning to distinguish between an
important “practical” difference and
“statistically” significant difference is a
valuable lesson for a Black Belt.

一元方差分析:
Analysis of Variance – One-way ANOVA
To introduce the concepts of Analysis of Variance
Sum of Squares
Mean Square Error
To demonstrate and practice calculating the ANOVA table
Manually
Minitab
To practice ANOVA
Exercises
Homework
ANOVA Introduction
Example:
A fertilizer company wants to compare fields treated with
four new, spring wheat fertilizers to fields with no fertilizer.
Ten different fields were sampled for each treatment and
the average bushels/acre was computed.
Problem:
Were any of the fertilizers different from each other and the
unfertilized control group?
Many times, there are more than two
groups of means. ANOVA solves the
problem of more than two means.
Multiple t-tests – Adequate?
An experimenter could run 2-sample t-tests against every
combination of means
1 vs. 2, 1 vs. 3, 1 vs. 4, 1 vs. 5
2 vs. 3, 2 vs. 4, 2 vs. 5
3 vs. 4, 3 vs. 5
4 vs. 5
Why would this not be a good idea?
Obviously, it is tedious and cumbersome
What about alpha risk?
Each t-test has a risk of false rejection ()